# Find the derivatives of the following [closed]

If $$y^4+5xy+y =2$$ find $$\frac{dy}{dx}$$ and d^2y/dx^2 when $$x=0$$ and $$y=1$$

I could not differentiate w.r.t x...answer coming to 0 which is not the correct answer.

$$y^4+5xy+y=2$$

$$\implies \frac d{dx} (y^4+5xy+y)=0$$

$$\implies 4y^3 \frac {dy}{dx}+5(y+x\frac {dy}{dx})+\frac {dy}{dx}=0$$

$$\implies (4y^3+5x+1)\frac {dy}{dx}+5y=0$$

Putting the values of $$x$$ & $$y$$

$$\implies 5\frac {dy}{dx}+5=0$$

$$\implies \frac {dy}{dx}=-1$$

We have $$4y^3 \frac{dy}{dx}+5y+5x\frac{dy}{dx}+\frac{dy}{dx}=0$$ and plugging $$x=0$$ and $$y=1$$ we get $$4\frac{dy}{dx}+5+\frac{dy}{dx}=0$$ and thus $$\frac{dy}{dx}=-1$$

Alternatively, refer to the inverse function: $$y^4+5xy+y =2 \iff x=\frac{2-y-y^4}{5y}=\frac2{5y}-\frac15-\frac{y^3}{5}$$ Hence: $$\frac{dy}{dx}|_{x=0}=\frac1{\frac{dx}{dy}|_{y=1}}=\frac1{\left(-\frac25-\frac{3y^2}{5}\right)|_{y=1}}=\frac1{-1}=-1.$$